Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{y - 1}{y^2 + 6y - 27} \times \dfrac{-3y - 27}{y - 1} $
Solution: First factor the quadratic. $z = \dfrac{y - 1}{(y + 9)(y - 3)} \times \dfrac{-3y - 27}{y - 1} $ Then factor out any other terms. $z = \dfrac{y - 1}{(y + 9)(y - 3)} \times \dfrac{-3(y + 9)}{y - 1} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (y - 1) \times -3(y + 9) } { (y + 9)(y - 3) \times (y - 1) } $ $z = \dfrac{ -3(y - 1)(y + 9)}{ (y + 9)(y - 3)(y - 1)} $ Notice that $(y - 1)$ and $(y + 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -3(y - 1)\cancel{(y + 9)}}{ \cancel{(y + 9)}(y - 3)(y - 1)} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $z = \dfrac{ -3\cancel{(y - 1)}\cancel{(y + 9)}}{ \cancel{(y + 9)}(y - 3)\cancel{(y - 1)}} $ We are dividing by $y - 1$ , so $y - 1 \neq 0$ Therefore, $y \neq 1$ $z = \dfrac{-3}{y - 3} ; \space y \neq -9 ; \space y \neq 1 $